Consider a right angled triangle with an acute angle of $\theta$.  Let the base of the triangle be of length 2.

Find the height of the triangle in terms of $t$, where $t=\tan \theta$.

Now imagine a line in the triangle which forms an isosceles triangle with two angles equal to $\theta$.

Use this diagram to prove the double angle formula, where $t=\tan \theta$: $$\tan2\theta = {2t\over {1-t^2}}, \quad \sin2\theta = {2t\over {1+t^2}},\quad \cos2\theta = {{1-t^2}\over {1+t^2}}$$